Question

Find the value of the least positive integer n for which ${{\left( \dfrac{1+i\sqrt{3}}{1-i\sqrt{3}} \right)}^{n}}=1$ .

Answer 1

Hint: First solve the term inside the bracket and then work on keeping n value one by one to find the least value which satisfies the condition required by question.

Complete step-by-step solution -
Definition of i, can be written as:
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
Given expression of n in the question is in the form:
${{\left[ \dfrac{\left( 1+i\sqrt{3} \right)}{\left( 1-i\sqrt{3} \right)} \right]}^{n}}=1$
Take the term inside the bracket outside separately and if we find its value in a simple form then we can substitute n to make equation true inside term:
$\dfrac{1+i\sqrt{3}}{1-i\sqrt{3}}$
Rationalize the term by multiplying and dividing with $1+i\sqrt{3}$ .By rationalizing the term, we get:
$\dfrac{\left( 1+i\sqrt{3} \right)\left( 1+i\sqrt{3} \right)}{\left( 1-i\sqrt{3} \right)\left( 1+i\sqrt{3} \right)}$
By using distributive law from concept of algebra:
a(b + c) = a.b + a.c
By using the distributive law our expression turns into the:
$\dfrac{1+i\sqrt{3}+i\sqrt{3}+i.i.3}{1+i\sqrt{3}-i\sqrt{3}-i.i.3}$
By simplifying the above expression our term turns into the:
$\dfrac{-2+2i\sqrt{3}}{4}$
By cancelling 4 with 2 in numerator our term turns into:
$\dfrac{-1+i\sqrt{3}}{2}=-\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}$
By using Euler’s formula of sin, cos of particular angle:
$e^{ix}=\cos x+i\sin x$
So, $-\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}=\sin \left( \dfrac{-\pi }{6} \right)+i\cos \left( \dfrac{-\pi }{6} \right)$
By substituting this into our expression our equation takes form of:
$\left( \sin \left( \dfrac{-n\pi }{6} \right)+i\cos \left( \dfrac{-n\pi }{6} \right) \right)=1$
Smallest n for which this is possible is 9.
There the least integer n for which given expression is true will be 9.

Note: The values 1, 2, 3,…..8 will not give 2 as quotient when divided by 6, by 9 gives it with the odd integer above. So, 9 is the least. This is a careful step, don't think wrong.